# Gas Cooling: Part 5

This is the last post in the series examining the science of gas cooling as a fire control tactic. Be forewarned, there is math ahead! I have made an attempt at providing sufficient explanation to allow firefighters, fire officers, and instructors to develop an understanding the scientific concepts underlying this fire control technique. My next post will return to the topic of extreme fire behavior and ventilation with discussion of the most recently released NIOSH report, Death in the Line of Duty 2010-10.

### The Mathematical Explanation

Dr. Stefan Särdqvist provides a mathematical explanation of volume changes during smoke/gas cooling In Water and Other Extinguishing Agents (Särdqvist,2002). Stefan’s text includes a graph that illustrates volume changes based on the extent to which the upper layer is cooled and the percentage of the water that is vaporized in the hot gases versus on contact with hot surfaces. As illustrated in Figure 1, the relative volume (expansion or contraction) of the upper layer during gas cooling is dependent on the percentage of water vaporizing as water passes through the hot gases of the upper layer and the percentage of water vaporizing on contact with hot surfaces such as compartment linings.

Figure 1. Volume Changes During Gas Cooling

Note: Adapted from Water and Other Extinguishing Agents (p. 155), by Stefan Särdqvist, 2002, Karlstad, Sweden: Räddnings Verket. Copyright 2002 by Räddnings Verket

If 100% of water applied for cooling vaporizes in the upper layer, the total volume of the hot fire gases and steam in the upper layer will be 79% of the original volume of the hot gases alone. If approximately 30% of the cooling water vaporizes in the upper layer and 70% vaporizes on contact with hot surfaces such as compartment linings (e.g., ceiling, walls) the volume of the upper layer will remain the same. However, if less than 30% of the cooling water is vaporized in the upper layer and the remainder is vaporized on contact with hot surfaces, the volume of the upper layer (hot fire gases and steam) will increase.

Understanding why this is the case requires a good understanding of the ideal gas law and a willingness to work through the math. As Greg Gorbett and Jim Pharr observe in the math review chapter of Fire Dynamics (2010), “The term algebra inspires dread in many otherwise competent, confident people” (p. 16).

### Gas Cooling and the Ideal Gas Law

Gas Cooling: Part 4, examined the expansion ratio of steam using the Ideal Gas Law, providing a worked example to illustrate how to solve for the change in volume when water is vaporized to steam. As illustrated below, the Ideal Gas Law can also be used to determine relative influences of contraction of the upper layer and expansion of steam during gas cooling.

Before the application of water:

After the application of water:

Where:

P=Pressure (Pa)

V= Volume (m3)

T=Temperature (K)

n=Moles

Ru=Universal Gas Constant (8.3145 J/mol K)

Subscript of 1 refers to initial conditions where the upper layer consists of hot smoke and air

Subscript of 2 refers to conditions at (later) time where the upper layer consists not only to the hot smoke and air, but also to the water applied for cooling (the number of molecules in the upper layer increases, and temperature changes).

Another way of expressing the initial and final conditions using the two gas laws is to set them equal to one another:

Pressure (P) in the fire compartment and adjacent compartments remains relatively constant (due to compartment openings and other leakage). For example, the National Fire Protection Association (NFPA) Standard 92A Standard for Smoke Control Systems Using Barriers and Pressure Differences (2009) specifies a design pressure difference of 24.9 Pa (0.0036 psi) to exclude smoke from a protected area (such as a stairwell) in a non-sprinklered building with 2.7 m (9’) ceilings. As atmospheric pressure is 102325 Pa (14.7 psi) the pressure difference, while significant enough to influence smoke movement is actually quite small in most cases. Given that pressure is relatively constant and the Universal Gas Constant (Ru) is the same for all ideal gases, these factors will have the same effect on initial and final conditions (allowing both Ru and P to be factored out of the ideal gas equations used to determine changes in upper layer volume.

After factoring out Ru and P, the relationship between the upper layer volume before and after application of water is as follows:

### Lots Going On!

When water is applied to cool the upper layer, there is quite a bit going on. Energy is transferred from the upper layer to the water, lowering the temperature of the upper layer and raising the temperature of the water to its boiling point, vaporizing the water, and raising the temperature of the resulting steam. As the absolute temperature of the upper layer is reduced, its volume is proportionally reduced. However, as water is vaporized at its boiling point and the absolute temperature of the resulting steam is increased its volume increases. The important question is where was the water vaporized? Water vaporized in the upper layer, absorbed energy from the hot gases, lowering their absolute temperature. However, water that passes through the hot gas layer and vaporizes on contact with hot surfaces such as compartment linings (e.g., ceiling, walls) did not absorb significant energy from the upper layer and did not significantly reduce the temperature of the upper layer. Steam produced as a result of water vaporizing on contact with hot surfaces can absorb energy from the upper layer, but this has far less impact than water vaporized within the upper layer due to the large difference between the specific heat of steam and the latent heat of vaporization of water.

While some energy is lost as a result of convection of hot gases out compartment openings and conduction through compartment linings and other structural materials, these factors are not considered in this analysis of the effect of gas cooling. In this analysis the compartment defines the bounds of the thermodynamic system and the gas cooling process is considered to be adiabatic (no energy is gained or lost by the system).

### Step by Step

I have made a few revisions to the explanation of gas cooling in Water and Other Extinguishing Agents (Särdqvist,2002), most significant of which is inclusion of the energy required to raise the temperature of the water applied for cooling to its boiling point (100o C). While the amount of energy is not large, this addition provides a more complete picture of the process involved in gas cooling. Other changes include consistent use of J/mol as units for specific heat and latent heat of vaporization, and minor variations in notation.

The mathematical explanation of gas cooling starts out in the same place as the concrete example provided in Gas Cooling Parts 1 and 2, determining the energy that must be transferred from the upper layer to water applied for cooling in order to achieve a specific reduction in temperature and the amount of water required to accomplish this. However, unlike an example using a specific compartment in which the units for specific heat and latent heat of vaporization were J/kg, the mathematical explanation uses J/mol (the reason for this will become clear as we dig a bit deeper).

The relationship between J/mol and kJ/kg as units of measure for specific heat and latent heat of vaporization is fairly straightforward as illustrated below:

The following equation explains the energy balance between hot gases in the upper layer and water applied for cooling. At first glance, this equation seems extremely complex, but if each segment is examined individually, it is fairly straightforward.

Where

Cp,g=Specific heat capacity of fire gases/smoke (approximately the same as air, 33.2 J/mol K at 1000 K).

Cp,st=Specific heat capacity of steam (41.2 J/mol K at 1000 K)

Cp,w=Specific heat capacity of water (76.663 J/mol K at 215.15 K)

LV,w=Latent heat of vaporization of water 40,680 J/mol

Tu=Temperature of the upper layer (K)

Tw=Temperature of water (K)

n=Moles

Subscript of 1 refers to initial conditions

Subscript of 2 refers to conditions at (later) time 2

First examine the left side of the equation which deals with the hot gases in the upper layer.

The left side of the equation determines the energy that must be transferred from the hot gases in the upper layer in order to result in a specific reduction in temperature. As this example does not deal with a specific compartment, the mass of the upper layer is unknown. A challenge resolved through the use of moles to define the amount of hot fire gases present in the upper layer. Remember that moles are a measure of the number of molecules present.

Multiplying the molar specific heat of smoke (Cp,g) in J/mol K by the number of moles (n) determines the energy that must be transferred from the upper layer to change its temperature 1 K. Multiplying that value by the change in absolute temperature (T1-T2) determines the total energy that must be transferred to achieve the specified change in absolute temperature.

Now examine the right side of the equation which deals with the water applied for cooling:

The right side of the equation determines the energy that must be transferred to the water applied for cooling in order to increase the temperature of the water (as steam) by the same extent as the reduction in upper layer temperature.

The first step is to determine the amount of water applied (remember the assumption that all water applied is vaporized either in the gas layer or on contact with surfaces). This is accomplished by subtracting the amount of hot gases in the upper layer (in Moles) from the amount of hot gases, and steam in the upper layer after cooling the gases (n2 – n1).

When vaporized in the upper layer, energy is transferred from the hot gases in the upper layer to 1) raise the temperature of the water to its boiling point of 373.15 K (100o C), 2) to change its state from liquid phase to gas phase, and 3) to raise the temperature of the steam until reaching equilibrium (hot gases and steam are at the same temperature). When water is vaporized on contact with a hot surface, it did not absorb significant energy while traveling through the hot gasses of the upper layer. The energy necessary to raise the temperature of the water to its boiling point and vaporize it is absorbed from the surface. Steam produced in this manner will also absorb energy from the hot gases of the upper layer (but the process of increasing the temperature of the water in liquid form and vaporization did not take significant energy from the hot gasses of the upper layer).

Figure 2. Gas Versus Surface Cooling

As water that is vaporized in the upper layer absorbs energy from the hot gases to raise its temperature to boiling and vaporize.

The temperature increase required for water to reach its boiling point is determined by subtracting the initial temperature of the water (Tw,1) from its boiling point of 373.15 K. The increase in the temperature of the water in liquid form is multiplied by the specific heat of water (Cp,w) to calculate the total energy required for this temperature increase (Cp,w (373.15- Tw,1)).

The latent heat of vaporization (LV,w) is added to the energy required to raise the temperature of the water from Tw1 to its boiling point of 373.15 K (100o C).

After water is vaporized (either while traveling through the upper layer or on contact with a hot surface) it continues to absorb energy from the upper layer until the temperature of the steam and the hot gases in the upper layer reach the same temperature and are in thermal equilibrium. The specific heat of steam (Cp,w) is multiplied by the difference between 373.15 K (100o C) and the final temperature of the upper layer (Tu2). This determines the energy required for the steam and the hot gases in the upper layer to reach thermal equilibrium ( ).

As with the calculations examining the smoke and hot gases in the upper layer, moles are a measure of the amount of water applied for cooling. As the thermodynamic system of the compartment is being treated as adiabatic (no energy leaves the system, it is simply transferred between the hot gases of the upper layer and the water applied for cooling), the left and right sides of the equation must be equal.

Solving for n, this equation may also be written:

Given that:

The left side of the equation can be simplified to solve for the amount of molecules in the upper layer before cooling (n1) and after application of water (n2) as follows:

Solving for n allows the energy exchange equation to be combined with the two ideal gas laws used to describe changes in volume associated with gas cooling.

As the energy exchange equation is equal to the initial amount of gas molecules in the upper layer before gas cooling divided by the amount of gas molecules in the upper layer after the application of water, the energy balance equation can be inserted in the ideal gas law in place of the amount of molecules (n1 and n2) as illustrated below:

This formula looks quite complex, but in actuality most of the values are constants such as the specific heat of water (Cp,w), latent heat of vaporization of water (LVw), and specific heat of steam (Cp,st). After plugging in these constants, the only variables on the right side of the equation are the temperature of the upper layer before and after cooling and the percentage of water vaporized in the upper layer.

The volume of the upper layer after cooling divided by the volume of the upper layer before cooling is the percentage change in volume of the upper layer.

### Worked Examples

While explaining the equations is important, there is nothing quite so useful in developing understanding as actual worked examples. In each of these examples, the initial upper layer temperature is 773.15 K (500o C), the initial temperature of the cooling water is 293.15 K (20o C) and the final temperature of the upper layer is 473.15 K (200o C).

Example 1: All (100%) of the water applied for cooling is vaporized in the upper layer.

In this example where all of the water applied for cooling is vaporized in the upper layer, the volume of the upper layer is reduced by 27% and the lower boundary of the upper layer would rise. This illustrates the ideal (but likely not achievable) application of gas cooling to reduce temperature and raise the lower boundary of the upper layer.

Example 2: None (0%) the water applied for cooling is vaporized in the upper layer; all of it is vaporized on contact with hot compartment linings or other surfaces.

In this example where none of the water applied for cooling is vaporized in the upper layer, but vaporizes on contact with hot surfaces, the volume of the upper layer would double. If the upper layer filled more than half the volume of the compartment, the upper layer would then fill the entire compartment with hot gases and steam at 473.15 K (200o C), providing an untenable environment for both firefighters and trapped occupants. This is why an indirect attack is not used from inside the compartment or in compartments where there may be savable victims.

Example 3: One third (33.3%) of the water applied for cooling vaporizes in the upper layer and the remainder (66.6%) is vaporized on contact with hot compartment linings or other surfaces.

In this example, the volume of the upper layer is unchanged, but considerably cooler than before the application of water. If firefighters had adequate working area below the upper layer before applying cooling water, this would be unchanged, but the temperature of the gases overhead would be considerably reduced. With good technique and an appropriate flow rate, more than 33.3% of the water applied for cooling can be vaporized in the upper layer, providing practical results somewhere between a 3% (Example 3) and 27% (Example 1) reduction in the volume of the upper layer. However, it is important to remember that the fireground is much more dynamic than the simple analysis presented in this post.

### Different Parts of the Elephant

Firefighter’s perspectives on the use of water fog for interior structural firefighting can be compared to the Indian fable of The Six Blind Men and the Elephant (Saxe, 1963). In this fable, the six men tried to determine what an elephant was. As none of the men could see, they used their sense of touch. However, each grasped a different part of the elephant. One touched the side and thought an elephant was like a wall, another the trunk and thought an elephant was like a snake, and so forth. What you believe may be limited by your point of observation.

Many firefighters in the United States find it hard to believe that the volume of the upper layer can be reduced and the bottom of the upper layer raised by application of water. This is inconsistent with their experiences in the field. The explanation provided in this post illustrates how this is possible. If flow rate and application technique used result in more than 70% of cooling water vaporizing on contact with hot surfaces, the upper layer will increase in volume and the level of the hot gas layer in a confined area such as a compartment will become lower (consistent with many firefighters experience when using water fog for interior structural firefighting). However, this does not have to be the case. Where the water is vaporized and the resulting effects are dependent on application technique, flow rate, and duration of application!

I would like to extend a great deal of thanks to Stefan Särdqvist for providing the basis for this explanation and to Lieutenant Felipe Baeza Lehnert of Valdivia (Chile) Fire Department Company 1 (Germania) for his patience in helping me sort though the math.

Ed Hartin, MS, EFO, MIFireE, CFO

### References

National Fire Protection Association (NFPA). (2009). Standard 92A Standard for Smoke Control Systems Using Barriers and Pressure Differences. Quincy, MA: Author.

Saxe, J. (1963). The blind men and the elephant. New York: McGraw-Hill

Särdqvist, S. (2002) Water and other extinguishing agents. Karlstad, Sweden: Räddnings Verket

### 5 Responses to “Gas Cooling: Part 5”

1. brian Says:

this is a great paper!It let me learned a lot!Thanks for your share!
And here is a question that if “n1″ of the right equation below” Solving for n, this equation may also be written:…” should’t exist? It confused me.
thank you for your share again!

2. Martin CHAMBERT Says:

Hi,thank you for these themodynamical lessons.
Before anything, I’m French, so please excuse me for my bad english.
I’m in an engineering school, and have a lot of thermodynamic tuitions per week, and the content of your article requires more than a simple explanation of the PV=nRT formula.
Moreover, you can’t justify the quantity you quote. Thermodynamic you are giving here is called “basic thermodynamic”, because you use a lot of simplification. For example, you consider that all the water is vaporized at the same time, that the gaz mix is homogeneous, that all the volume of the smoke layer undergoes the cooling in a constant way.
So between theories and reality, there is a huge gap. At least considering firefighting.
In industries it’s different because you use systems that provides to approach such considerations.
When you blow a water stream, the effect is firstly local, and then, there is a thermodynamic balance which happens.
But this takes a time, and has effects on smoke movement (speed, direction…)

It’s important to learn to others what is going on when you open the nozzle, but it’s almost as important to give a comparison between theories and action. And I don’t speak about labo studies. I mean on real fires (FDS).

But I appreciate the way you are giving yourself such difficulties to explain what is going underneath.

Just an example if you think I’m wrong. Theorically, I don’t want to give any precise volumes, but let take 50 liters of water. Imagine it’s the result of your calculations. Take a 1m x 50cm *10cm box, full of liquid water, you agree the cooling won’t be the same if you put those 50 liters in this shape in the middle of your smoke layer, or if you throw it with a nozzle. But theorically it is!!!

Thanks for the work you’ve done. If you want to contact me, don’t hesitate.
Bye
Martin CHAMBERT

3. hartin Says:

Martin, You are absolutely correct. There is a tremendous difference between the simple explanation and real world firefighting operations. As I quoted in an earlier post: “All models are wrong, but some are useful” (Box & Draper). Each of the elements of this explanation is simplified to illustrate the basic concepts, but is considerably more complex in reality. Cheers, Ed

4. hartin Says:

Brian, I am not sure that I understand your question, but I will review the post. Cheers, Ed

5. brian Says: